Question

Which of the following reactions would give caproic acid?
A.$n - {C_5}{H_{11}}Br\xrightarrow[{(ii)C{O_2}}]{{(i)Mg/Ether}}$
B.$n - {C_5}{H_{11}}Li\xrightarrow[{(ii){H_3}{O^ + }}]{{(i)C{O_2}}}$
C.$n - {C_4}{H_9}Br\xrightarrow[{(ii)C{O_2}(iii){H_3}{O^ + }}]{{(i)Mg/Ether}}$
D.$n - {C_5}{H_{11}}MgBr + {(CN)_2}\xrightarrow[{(ii){H_3}{O^ + }}]{{(i)\Delta }}$

Answer 1

The term ‘caproic acid’ is the common name given to hexanoic acid. It is a member of the carboxylic acid group. As the name suggests, it has six carbon atoms. Its molecular formula is ${C_5}{H_{11}}COOH$ . It can be prepared by any method that can be applied for the production of carboxylic acids.

Complete step by step answer:
A.When we react n-bromopentane ($n - {C_5}{H_{11}}Br$) with magnesium metal in the presence of dry ether and carbon dioxide, then we get caproic acid (i.e., hexanoic acid). The reaction can be given as follows:
$n - {C_5}{H_{11}}Br\xrightarrow[{(ii)C{O_2}}]{{(i)Mg/Ether}}{C_5}{H_{11}}COOH$
B.When we react $n - {C_5}{H_{11}}Li$ with carbon dioxide followed by hydrolysis, we obtain the following product:
$n - {C_5}{H_{11}}Li\xrightarrow[{(ii){H_3}{O^ + }}]{{(i)C{O_2}}}{C_5}{H_{11}} - CO - {C_5}{H_{11}}$
C.When we react n-bromobutane ($n - {C_4}{H_9}Br$) with magnesium metal in the presence of dry ether and then with carbon dioxide followed by hydrolysis, then we get pentanoic acid. The chemical reaction can be given as follows:
$n - {C_4}{H_9}Br\xrightarrow[{(ii)C{O_2}(iii){H_3}{O^ + }}]{{(i)Mg/Ether}}{C_4}{H_9}COOH$
D.When we heat pentyl magnesium bromide ($n - {C_5}{H_{11}}MgBr$) with ${(CN)_2}$followed by hydrolysis, then we get caproic acid as follows:
$n - {C_5}{H_{11}}MgBr + {(CN)_2}\xrightarrow{{(i)\Delta }}{C_5}{H_{11}}CN + MgBrCN\xrightarrow{{(ii){H_3}{O^ + }}}{C_5}{H_{11}}COOH$
Thus option-A and D are correct.

Note:
The formation of caproic acid takes place as follows:
-When an alkyl halide reacts with magnesium in the presence of dry ether, we obtain a Grignard reagent.
-When a Grignard reagent reacts with carbon dioxide followed by the addition of a proton, then a carboxylic acid is formed with one carbon more than the Grignard reagent.
-Hence we obtain hexanoic acid from n-bromopentane.
-Also, when we react to a Grignard reagent with cyanogen ${(CN)_2}$, then we get hexanenitrile.
-Hexanenitrile gives hexanoic acid upon further hydrolysis.